Question

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3.

Answer 1

HELLO DEAR,

A sphere of radius (r) is given.
Let R and h be the radius and the height of the cone respectively.
and the volume of the cone is V = 1/3πR²h


from right ∆BCD,

BC² = r² - R²

h = r + BC²

V = 1/3πR²{r + √(r² - R²)}
= 1/3πR²r + 1/3πR²√(r² - R²)
now,
dV/dR = 2/3πRr + 2/3πR√(r² - R²) + πR²/3 * (-2R)/2√r² - R²

= 2/3πRr + 2/3πR√(r² - R²) - πR³/3√(r² - R²)

= 2/3πRr + $\bold{\frac{2\pi R(r^2 - R^2) - \pi R^3}{3\sqrt{r^2 - R^2}}}$

= 2/3πRr + $\bold{\frac{2\pi Rr^2- 3\pi R^3}{3\sqrt{r^2 - R^2}}}$

now, dV/dR = 0

2/3πRr = [3πR³ - 2πRr²]/3√(r² - R²)

2r√(r² - R²) = 3R² - 2r²
now on squaring both side,

4r²(r² - R²) = 9R⁴ + 4r⁴ - 12R²r²

4r⁴ - 4r²R² - 4r⁴ + 12r²R² = 9R⁴

8r²R² = 9R⁴

R² = 8r²/9



Now, d²V/dR² = 2/3πr + $\bold{\frac{3\sqrt{r^2-R^2}(2\pi r^2 - 9\pi R^2) - (2\pi Rr^2 - 3\pi R^3)(-6R)\frac{1}{2\sqrt{r^2 - R^2}}}{9(r^2 - R^2)}}$

d²V/dR² = 2/3πr + $\bold{\frac{3\sqrt{r^2-R^2}(2\pi r^2 - 9\pi R^2) + (2\pi Rr^2 - 3\pi R^3)(3R)\frac{1}{\sqrt{r^2 - R^2}}}{9(r^2 - R^2)}}$

when, R² = 8r²/9,
it can shown that d²V/dR² < 0.

therefore, volume is maximum when

R² = 8r²/9

so, height of the cone h = r + √(r² - R²)

h = $\bold{r + \sqrt{r^2 - \frac{8r^2}{9}}}$

h = $\bold{r + \sqrt{r^2/9}}$

h = $\bold{r + r/3}$

h = 4r/3

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3.

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Step-by-step explanation:

I hope this would be a correct answer